Catch That Cow

Catch That Cow

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice POJ 3278

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;




const int maxn=100001;
bool a[maxn];
int step[maxn];
queue <int> q;
int bfs(int n,int k)
{
    int head,next;
    q.push(n);
    step[n]=0;
    a[n]=1;
    while(!q.empty())
    {
        head=q.front();
        q.pop();
        for(int i=0; i<3; i++)
        {
            if(i==0) next=head-1;
            else if(i==1) next=head+1;
            else next=head*2;
            if(next<0||next>=maxn) continue;
            while(!a[next])
            {
                a[next]=1;
                step[next]=step[head]+1;
                q.push(next);

            }
            if(next==k) return step[next];
        }
    }
}
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        memset(a,0,sizeof(a));
        memset(step,0,sizeof(step));
        if(n>k) printf("%d\n",n-k);
        else printf("%d\n",bfs(n,k));
    }
    return 0;
}